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Proof step 1 from proof pls (conclusion)

m < m0, n < n0 ⊦ m + n + 1 < m0 + n0

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ m < m0 ∧ n < n0 → (¬ m + n < m0 + n0 ↔ false)
Proof step 3:
m < m0 ∧ n + 1 < n0 → (¬ m + n + 1 < m0 + n0 ↔ false), m < m0, n < n0 ⊦ m + n + 1 < m0 + n0

Rule argument

spec: nat-basic2 inst: name: l-03 seq: ⊦ m < m0 ∧ n < n0 → (¬ m + n < m0 + n0 ↔ false) subst: [m, m0, n, n0] → [m, m0, n + 1, n0] flags: current, []

Additional information

Used terms: m, m0, n + 1, n0